MHT-CET Maths · Determinants and Matrices

Determinants, Cofactors and the Adjoint Identities

Cofactors build both the determinant and the adjoint — and three identities, A·adj(A) = |A|I, |adj A| = |A|ⁿ⁻¹ and |kA| = kⁿ|A|, answer most of what MHT-CET asks about them.

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Why this matters

16 PYQs at 69% HARD — the hardest page in the chapter and, at the same time, its most learnable, because the HARD questions are recalled identities in disguise rather than long computations. 'Find α given adj A and |A|' has been set three times, 'A·adj A = AAᵀ, find a and b' four times in three sittings, and the cofactor expansion appears both as a matrix of cofactors and as a single element of the adjoint. Learn the three identities as facts; the page then costs about a minute a question.

Concept 1 of 5

Determinants and Cofactors: Expansion Along a Row

Intuition

A 3 × 3 determinant is a signed sum of 2 × 2 determinants: each entry of one row times its cofactor. The cofactor is the minor with a checkerboard sign attached, and the same cofactors are the raw material of the adjoint.

Definition

$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ . For $3 \times 3$ : expand along any row or column, $|A| = \sum_j a_{ij}A_{ij}$ , where $A_{ij} = (-1)^{i+j}M_{ij}$ and $M_{ij}$ is the minor (delete row $i$ , column $j$ ).
Sign checkerboard: $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ .
Expanding along a row with its OWN cofactors gives $|A|$ ; with another row's cofactors (an 'alien' expansion) it gives $0$ . So $a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = |A|$ — for a rotation-type matrix that is $\cos^2\theta + \sin^2\theta = 1$ .
The matrix of cofactors $[A_{ij}]$ is computed entry by entry; the adjoint is its transpose, so $(\operatorname{adj}A)_{23} = A_{32}$ , the cofactor of the entry in row 3, column 2.
A determinant with a parameter ( $x, y, z$ on the diagonal) expands to a polynomial; substitute the given relations at the end, not the beginning.

Expansion and cofactors

|A| = \sum_{j} a_{ij}A_{ij}, \qquad A_{ij} = (-1)^{i+j}M_{ij}, \qquad \sum_j a_{ij}A_{kj} = 0 \ (k \ne i)

Worked example

For

A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 1 & 1 \\ 0 & 2 & 4 \end{pmatrix}

, find the cofactor

A_{23}

and the element

(\operatorname{adj}A)_{32}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Determinants and MatricesMODERATE

A=\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ - \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}

, where

A_{21},A_{22},A_{23}

are cofactors of

a_{21},a_{22},a_{23}

respectively, then the value of

a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}=

[Q149 · 19 April Shift I · 2025]

Reading (adj A)₂₃ as the cofactor A₂₃

The adjoint is the TRANSPOSE of the cofactor matrix, so its

(2, 3)

element is

A_{32}

. With

a_{ij} = 2i + j

that gives

4

, and the untransposed reading gives

-4

— both on the option list.

Drill 2 more on determinants and cofactors: expansion along a row

Concept 2 of 5

The Adjoint and A·adj(A) = |A|·I

Intuition

Multiplying

A

by its adjoint puts each row's own cofactor expansion on the diagonal (giving

|A|

) and alien expansions off it (giving

0

). So

A\cdot\operatorname{adj}A

|A|

times the identity — which is why the inverse is

\operatorname{adj}A/|A|

Definition

$A\,\operatorname{adj}A = \operatorname{adj}A\,A = |A|\,I$ . If $A\,\operatorname{adj}A = 20I$ , then $|A| = 20$ — read it off the diagonal.
For $2 \times 2$ : $\operatorname{adj}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ — swap the diagonal, negate the off-diagonal.
$\operatorname{adj}A = |A|\,A^{-1}$ , hence $(\operatorname{adj}A)^{-1} = \dfrac{A}{|A|}$ .
When $|A|$ is given through relations ( $xyz = 60$ , $8x + 4y + 3z = 20$ ), expand $|A|$ symbolically first — $xyz - 8x - 4y - 3z + 28$ — then substitute: $60 - 20 + 28 = 68$ , so $A\,\operatorname{adj}A = 68I$ .
A polynomial in $\operatorname{adj}A$ with binomial coefficients, $I - 3M + 3M^2 - M^3$ , is $(I - M)^3$ : compute $I - M$ once and cube it.

The adjoint identity

A\,\operatorname{adj}A = |A|\,I \qquad \operatorname{adj}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \qquad (\operatorname{adj}A)^{-1} = \frac{A}{|A|}

Worked example

For

A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}

, compute

A\,\operatorname{adj}A

without multiplying the matrices.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Determinants and MatricesEASY

For an invertible matrix

A

, if

A(\text{adj}\,A) = \begin{pmatrix}20 & 0\\0 & 20\end{pmatrix}

, then

|A| =

[Q116 · May Shift 1 · 2021]

Substituting the relations before expanding

|A| = xyz - 8x - 4y - 3z + 28

contains BOTH given quantities and a constant. Using only

8x + 4y + 3z = 20

gives

20I

; the full expansion gives

68I

. Expand first, substitute last.

Drill 3 more on the adjoint and a·adj(a) = |a|·i

Concept 3 of 5

|adj A| = |A|ⁿ⁻¹ and |kA| = kⁿ|A|

Intuition

Take determinants of

A\,\operatorname{adj}A = |A|\,I

: the right side is

|A|^n

, so

|\operatorname{adj}A| = |A|^{n-1}

. A scalar multiplies EVERY row, so it comes out of the determinant once per row:

|kA| = k^n|A|

Definition

$|\operatorname{adj}A| = |A|^{n-1}$ : for $3 \times 3$ with $|A| = 4$ , $|\operatorname{adj}A| = 16$ ; with $|A| = 5$ , $25$ .
The recurring stem: ' $P$ is the adjoint of a $3 \times 3$ matrix $A$ with $|A| = 4$ ; find $\alpha$ '. Expand $|P|$ as a linear expression in $\alpha$ , set it equal to $|A|^2 = 16$ , solve.
$|kA| = k^n|A|$ , $|A^T| = |A|$ , $|AB| = |A||B|$ , $|A^{-1}| = \dfrac{1}{|A|}$ , $|A^m| = |A|^m$ .
$\operatorname{adj}(kA) = k^{n-1}\operatorname{adj}A$ and $\operatorname{adj}(\operatorname{adj}A) = |A|^{n-2}A$ — rarer, but the same family.

Determinant identities

|\operatorname{adj}A| = |A|^{n-1} \qquad |kA| = k^n|A| \qquad |AB| = |A||B| \qquad |A^{-1}| = \frac{1}{|A|}

Worked example

Q = \begin{pmatrix} 2 & 0 & 1 \\ 1 & \beta & 0 \\ 0 & 1 & 1 \end{pmatrix}

is the adjoint of a

3 \times 3

matrix

A

with

|A| = 3

. Find

\beta

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Determinants and MatricesHARD

P = \begin{pmatrix}1&\alpha&3\\1&3&3\\2&4&4\end{pmatrix}

is the adjoint of a

3\times3

matrix A and

|A|=4

, then value of

\alpha

[Q127 · 11th May Shift 1 · 2024]

Using |adj A| = |A|

For

3 \times 3

|\operatorname{adj}A| = |A|^2

. Setting

|P| = 4

instead of

16

gives

\alpha = 5

, which is the option planted for that mistake.

Drill 2 more on |adj a| = |a|ⁿ⁻¹ and |ka| = kⁿ|a|

Concept 4 of 5

A·adj(A) = AAᵀ: Two Equations From the Diagonal and Off-Diagonal

Intuition

A\,\operatorname{adj}A

|A|

times the identity, so setting it equal to

AA^T

forces

AA^T

to be diagonal with

|A|

on the diagonal. The off-diagonal entry of

AA^T

must be

0

and each diagonal entry must equal

|A|

— two equations in the two unknowns.

Definition

Write $A = \begin{pmatrix} p & q \\ 3 & 2 \end{pmatrix}$ with $p, q$ in terms of the unknowns. $AA^T = \begin{pmatrix} p^2 + q^2 & 3p + 2q \\ 3p + 2q & 13 \end{pmatrix}$ , and $|A| = 2p - 3q$ .
Off-diagonal: $3p + 2q = 0$ . Diagonal: $13 = |A| = 2p - 3q$ . Solve the pair.
For $A = \begin{pmatrix} 2a & -3b \\ 3 & 2 \end{pmatrix}$ : $6a - 6b = 0$ and $4a + 9b = 13$ give $a = b = 1$ , so $2a + 3b = 5$ . For $\begin{pmatrix} 5a & -b \\ 3 & 2 \end{pmatrix}$ : $15a - 2b = 0$ , $10a + 3b = 13$ give $a = \frac25$ , $b = 3$ , so $5a + b = 5$ .
The other diagonal entry $p^2 + q^2 = |A|$ is automatically satisfied once the first two hold — use it as a check, not a third equation.

The AAᵀ condition

A\,\operatorname{adj}A = AA^T \iff AA^T = |A|\,I \iff (\text{off-diagonal of } AA^T) = 0 \ \text{and}\ (\text{diagonal of } AA^T) = |A|

Worked example

A = \begin{pmatrix} 3a & -2b \\ 2 & 3 \end{pmatrix}

and

A\,\operatorname{adj}A = AA^T

, find

a

and

b

Practice this conceptself-check

From the bank · past-year question

Example 4Determinants and MatricesHARD

A = \begin{pmatrix}2a & -3b\\3 & 2\end{pmatrix}

and A·adj A = AA

^T

, then

2a+3b

[Q139 · 13th May Shift 1 · 2024]

Equating AAᵀ to |A| only on the diagonal

The off-diagonal condition is the one that links

a

and

b

. Without it the diagonal gives one equation in two unknowns and every option looks reachable.

Drill 3 more on a·adj(a) = aaᵀ: two equations from the diagonal and off-diagonal

Concept 5 of 5

Determinant Equations: When Does |A| Vanish?

Intuition

A determinant set to zero is a condition on its entries. Expand, simplify with identities, and read the condition — a trigonometric determinant collapses to

1 + \cos 2B = 0

, and a matrix over cube roots of unity is singular exactly when one factor

(1 - a\omega)

vanishes.

Definition

Expand fully before simplifying: $\begin{vmatrix} \cos(A+B) & -\sin(A+B) & \cos 2B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B \end{vmatrix}$ along row 1 gives $\cos^2(A+B) + \sin^2(A+B) + \cos 2B = 1 + \cos 2B$ . Zero means $\cos 2B = -1$ , $B = (2n+1)\dfrac{\pi}{2}$ .
Over cube roots of unity ( $\omega^3 = 1$ , $1 + \omega + \omega^2 = 0$ ): $\begin{vmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{vmatrix} = (1 - a\omega)(1 - c\omega)$ , so the matrix is non-singular unless $a = \omega^2$ or $c = \omega^2$ . With $a, b, c \in \{\omega, \omega^2\}$ : $a = c = \omega$ forced, $b$ free — $2$ matrices.
Singular means $|A| = 0$ : no inverse, and $AX = 0$ has non-trivial solutions (the linear-systems page).
Count carefully: a 'number of distinct matrices' stem is a product of the free choices after the determinant condition has removed the bad ones.

Vanishing determinant

|A| = 0 \iff A \text{ singular} \iff A^{-1} \text{ does not exist} \qquad 1 + \omega + \omega^2 = 0,\ \omega^3 = 1

Worked example

For what values of

\theta

\begin{vmatrix} \cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} = 0

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Determinants and MatricesHARD

\begin{vmatrix}\cos(A+B) & -\sin(A+B) & \cos 2B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{vmatrix} = 0

, then the value of

B

[Q129 · 9th May Shift 2 · 2024]

Stopping at cos 2B = 0

The trigonometric determinant is

1 + \cos 2B

, not

\cos 2B

: the two squared terms add to

1

. Zero requires

\cos 2B = -1

, giving

(2n+1)\pi/2

— the option

(2n+1)\pi/4

is the half-finished version.

Drill 1 more on determinant equations: when does |a| vanish?

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Determinants and Cofactors: Expansion Along a Row
Expansion and cofactors
$|A| = \sum_{j} a_{ij}A_{ij}, \qquad A_{ij} = (-1)^{i+j}M_{ij}, \qquad \sum_j a_{ij}A_{kj} = 0 \ (k \ne i)$
The Adjoint and A·adj(A) = |A|·I
The adjoint identity
$A\,\operatorname{adj}A = |A|\,I \qquad \operatorname{adj}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \qquad (\operatorname{adj}A)^{-1} = \frac{A}{|A|}$
|adj A| = |A|ⁿ⁻¹ and |kA| = kⁿ|A|
Determinant identities
$|\operatorname{adj}A| = |A|^{n-1} \qquad |kA| = k^n|A| \qquad |AB| = |A||B| \qquad |A^{-1}| = \frac{1}{|A|}$
A·adj(A) = AAᵀ: Two Equations From the Diagonal and Off-Diagonal
The AAᵀ condition
$A\,\operatorname{adj}A = AA^T \iff AA^T = |A|\,I \iff (\text{off-diagonal of } AA^T) = 0 \ \text{and}\ (\text{diagonal of } AA^T) = |A|$
Determinant Equations: When Does |A| Vanish?
Vanishing determinant
$|A| = 0 \iff A \text{ singular} \iff A^{-1} \text{ does not exist} \qquad 1 + \omega + \omega^2 = 0,\ \omega^3 = 1$

Watch out for (5)

Reading (adj A)₂₃ as the cofactor A₂₃
→ Determinants and Cofactors: Expansion Along a Row
Substituting the relations before expanding
→ The Adjoint and A·adj(A) = |A|·I
Using |adj A| = |A|
→ |adj A| = |A|ⁿ⁻¹ and |kA| = kⁿ|A|
Equating AAᵀ to |A| only on the diagonal
→ A·adj(A) = AAᵀ: Two Equations From the Diagonal and Off-Diagonal
Stopping at cos 2B = 0
→ Determinant Equations: When Does |A| Vanish?

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